Question

How many ways are there to select\(12\)countries in the United Nations to serve on a council if\(3\)are selected from a block of\(45\),\(4\)are selected from a block of\(57\), and the others are selected from the remaining\(69\)countries?

Short answer

The required ways are\(62,994,022,035,644,700\).

Step-by-step solution

Definitions of Product rule, Permutation, and Combination

Product rule: If one event can occur in\(m\)ways and a second event can occur in\(n\)ways, then the number of ways that the two events can occur in sequence is then\(m \cdot n\).

Definition permutation (order is important):

\(P(n,r) = \frac{{n!}}{{(n - r)!}}\)

Definition combination (order is not important):

\(C(n,r) = \left( {\begin{array}{*{20}{l}}n\\r\end{array}} \right) = \frac{{n!}}{{r!(n - r)!}}\)with\(n! = n \cdot (n - 1) \cdot \ldots \cdot 2 \cdot 1\)

Calculate the number of ways by the formula of Product rule, Permutation or Combination

\(3\)are selected from the block of\(45\):

\(\begin{array}{l}C(45,3) = \frac{{45!}}{{3!(45 - 3)!}}\\C(45,3) = \frac{{45!}}{{3!42!}}\\C(45,3) = 14,190\end{array}\)

\(4\)are selected from the block of\(57\):

\(\begin{array}{l}C(57,4) = \frac{{57!}}{{4!(57 - 4)!}}\\C(57,4) = \frac{{57!}}{{4!53!}}\\C(57,4) = 395,010\end{array}\)

\(12 - 4 - 3 = 5\)are selected from the block of\(69\)(since in total,\(12\)countries are selected)

\(\begin{array}{l}C(69,5) = \frac{{69!}}{{5!(69 - 5)!}}\\C(69,5) = \frac{{69!}}{{5!64!}}\\C(69,5) = 11,238,513\end{array}\)

Use the product rule:

\(14,190 \cdot 395,010 \cdot 11,238,513 = 62,994,022,035,644,700\)

Hence, the required ways are\(62,994,022,035,644,700\).